Download App

3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
$\left| {\,\begin{array}{*{20}{c}}{1/a}&{{a^2}}&{bc}\\{1/b}&{{b^2}}&{ca}\\{1/c}&{{c^2}}&{ab}\end{array}\,} \right| = $
  • A
    $abc$
  • B
    $1/abc$
  • C
    $ab + bc + ca$
  • ✓
    $0$
✓

Answer

Correct option: D.
$0$
d
(d)$\left| {\,\begin{array}{*{20}{c}}{1/a}&{{a^2}}&{bc}\\{1/b}&{{b^2}}&{ca}\\{1/c}&{{c^2}}&{ab}\end{array}\,} \right|$$ = \frac{1}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}1&{{a^3}}&{abc}\\1&{{b^3}}&{abc}\\1&{{c^3}}&{abc}\end{array}\,} \right| = \frac{{abc}}{{abc}}\left| {\,\begin{array}{*{20}{c}}1&{{a^3}}&1\\1&{{b^3}}&1\\1&{{c^3}}&1\end{array}\,} \right| = 0$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 3 and 4 . determinant and metrices→3 and 4 . determinant and metrices — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If ${x^m}{y^n} = {(x + y)^{m + n}}$ then ${\left. {{{dy} \over {dx}}} \right|_{x = 1,y = 2}}$ is equal to
→
Rolle's theorem is true for the function $f(x) = {x^2} - 4 $ in the interval
→
If $\theta=\tan ^{-1} a, \phi=\tan ^{-1} b$ and $a b=-1$, then $|\theta-\phi|$ is equal to
→
Let $f$ be a twice differentiable function on $R$. If $f^{\prime}(0)=4$ and $f(x)+\int_{0}^{x}(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$ then $(2 a+1)^{5} a^{2}$ is equal to $\dots\dots$
→
Let : $\overrightarrow{ a }=\hat{ i }+2 \hat{ j }+3 \hat{ k }, \overrightarrow{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }$ and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$ be there vectors. If $\vec{r}$ is a vector such that, $\overrightarrow{ r } \times \overrightarrow{ b }=\overrightarrow{ c } \times \overrightarrow{ b }$ and $\overrightarrow{ r } \cdot \overrightarrow{ a }=0$. Then $25|\overrightarrow{ r }|^2$ is equal to
→
In a $\triangle\text{ABC},$ if C is a right angle, then

$\tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{b}}\Big)=$

  1. $\frac{\pi}{3}$

  2. $\frac{\pi}{4}$

  3. $\frac{5\pi}{2}$

  4. $\frac{\pi}{6}$

→
Let $A=\left[\begin{array}{l}a_{1} \\ a_{2}\end{array}\right]$ and $B=\left[\begin{array}{l}b_{1} \\ b_{2}\end{array}\right]$ be two $2 \times 1$ matrices with real entries such that $A = XB,$

where $X=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & -1 \\ 1 & k\end{array}\right],$ and $k \in R$. If $a _{1}^{2}+ a _{2}^{2}=\frac{2}{3}\left( b _{1}^{2}+ b _{2}^{2}\right)$ and $\left( k ^{2}+1\right) b _{2}^{2} \neq-2 b _{1} b _{2}$ then the value of $k$ is ....... .

→
The __________ is the method available for solving an L.P.P
→
The inverse of the matrix $\left[ {\begin{array}{*{20}{c}}3&{ - 2}\\1&4\end{array}} \right]$is
→
The altitude of a cone is 20cm and its semi-vertical angle is 30°. If the semi-vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of:
  1. $30\text{cm}/\text{sec}.$
  2. $\frac{160}{3}\text{cm}/\text{sec}.$
  3. $10\text{cm}/\text{sec}.$
  4. $160\text{cm}/\text{sec}.$
→